I wanted to know a specific thing about Matlab fsolve technique. As shown in the code, there are two non-linear equations. I can normally solve them but I want to optimise the process. It is observed that there is a common term 2*x(1)^2 - 3*x(2)^3 in both of these equations. The question is, can I declare a separate term for the common part and use the fsolve(as in the commented part)? If yes, can anyone please tell me the way to do.
function f = sym4(x)
f(1)= 8*x(1)^2 -12*x(1)^3 +x(1)*x(2)+ x(2)^2-1;
f(2)= 2*x(1)^3 - 3*x(1)^4 +x(2)-3;
% z = 2*x(1)^2 - 3*x(2)^3;
%
% f(1)= 4*z + x(2)^2 + x(1)*x(2) -1;
% f(2)= x*z +x(2)-3;
end
Yes, you can use intermediate variables, which may be somewhat faster because the common term is only evaluated once. As horchler pointed out, you should check your indexes.
Related
Dear community of stack overflow. Is it possible to improve the speed of this code in Matlab? Can I use vectorization? Note that I have to use in every loop the "vpasolve" or "fsolve" function.
Here is the code which calls the function with a double loop:
for i=1:1000
for j=1:1000
SilA(i,j)=SolW(i,j);
end
end
Here is the function which is called by the above code. Can I vectorize the input of w, xi and make the code run faster?
function [ffw] = SolW(w,xi)
format long e;
z=0;mm=0.46;sop=80;
epit=0.1;voP=80.;
rho=2.1;aposC=0.1;aposD=0.1;
parEh=0.2*10^6;parEv=0.2*10^6;parGv=0.074*10^6;
parpos=0.35;hp=0.2;Ep=30*10^6;
parposV=0.20;ll=0.15;dd=2*ll;
C11=(parEh*(parEv-parEh*parpos^2)/((1+parpos)*(parEv-parEv*parpos-2*parEh*parpos^2)))*(1+2*1i*aposD);
C13=(parEh*parEv*parpos/((parEv-parEv*parpos-2*parEh*parpos^2)))*(1+2*1i*aposD);
C33=((1-parpos)*parEv^2/(parEv-parEv*parpos-2*parEh*parpos^2))*(1+2*1i*aposD);
C44=parGv*(1+2*1i*aposD);
DD=(Ep*hp^3)/(12*(1-parposV^2));
a1=C44;
a2=C33;
c1=(C13+C44)*1i*xi;
c2=(C13+C44)*1i*xi;
b1=rho*w^2-C11*xi^2;
b2=rho*w^2-C44*xi^2;
syms xr
rsol=vpasolve((a1*xr+b1).*(a2*xr+b2)-c1*c2*xr==0,xr);
rsol=eval(rsol);
r=[-sqrt(rsol)];
r1=r(1,:);
r2=r(2,:);
Fdf1=sop*sqrt(2*pi/(1i*epit*xi))*exp(1i*(xi*voP+w)^2/(2*epit*xi));
BC11=C44*(r1-1i*xi*c2*r1/(a2*r1^2+b2));
BC21=(C13*1i*xi)-((C33*c2*r1^2)/(a2*r1^2+b2))+(DD*xi^4-mm*w^2+1i*aposC*w)*c2*r1/(a2*r1^2+b2);
BC12=C44*(r2-1i*xi*c2*r2/(a2*r2^2+b2));
BC22=(C13*1i*xi)-((C33*c2*r2^2)/(a2*r2^2+b2))+(DD*xi^4-mm*w^2+1i*aposC*w)*c2*r2/(a2*r2^2+b2);
syms As1 As2;
try
[Ass1,Ass2]=vpasolve(BC11*As1+BC12*As2==0,BC21*As1+BC22*As2+Fdf1==0,As1,As2);
A1=eval(Ass1);
A2=eval(Ass2);
catch
A1=0.0;
A2=0.0;
end
Bn1=-(c2*r1/(a2*r1^2+b2))*A1;
Bn2=-(c2*r2/(a2*r2^2+b2))*A2;
ffw=Bn1*exp(r1*z)+Bn2*exp(r2*z);
end
Everything in your function but the calls to vpasolve, and try.... can be vectorize.
First, all this does not depend on w or xi, so could be computed once only:
format long e;
z=0;mm=0.46;sop=80;
epit=0.1;voP=80.;
rho=2.1;aposC=0.1;aposD=0.1;
parEh=0.2*10^6;parEv=0.2*10^6;parGv=0.074*10^6;
parpos=0.35;hp=0.2;Ep=30*10^6;
parposV=0.20;ll=0.15;dd=2*ll;
C11=(parEh*(parEv-parEh*parpos^2)/((1+parpos)*(parEv-parEv*parpos-2*parEh*parpos^2)))*(1+2*1i*aposD);
C13=(parEh*parEv*parpos/((parEv-parEv*parpos-2*parEh*parpos^2)))*(1+2*1i*aposD);
C33=((1-parpos)*parEv^2/(parEv-parEv*parpos-2*parEh*parpos^2))*(1+2*1i*aposD);
C44=parGv*(1+2*1i*aposD);
DD=(Ep*hp^3)/(12*(1-parposV^2));
a1=C44;
a2=C33;
From what I know, eval is slow, and I'm pretty sure that you don't need it:
rsol=eval(rsol);
Here is an example of vectorization. You should first generate all indices combination using meshgrid, and then use the . to noticed matlab to use element wise operations:
[I, J] = meshgrid(1:1000, 1:1000)
c1=(C13+C44)*1i.*xi;
c2=(C13+C44)*1i.*xi;
b1=rho.*w.^2 - C11.*xi.^2;
b2=rho.*w.^2-C44.*xi.^2;
But you won't be able to vectorize vpasolve, and try.... litteraly, without changing it to something else. vpasolve is probably the bottleneck of you computation (verify this using matlab profiler), so optimizing as proposed above will probably not reduce your computation time much.
Then you have several solutions:
use parfor if you have access to it to parallelize your computations, which depending on your architecture, could give you a 4x speedup or so.
it may be possible to linearize your equations and solve them all in one operation. Anyway, using a linear solver will be probably much faster than using vpasolve. This will probably give you the fastest speedup (guessing factor 10 -100 ?)
because you have continuous data, you could reduce the number of steps, if you dare loosing precision.
Hope this helps
In the above program everything can be vectorized as #beesleep said above.
For example:
[I, J] = meshgrid(1:1000, 1:1000)
c1=(C13+C44)*1i.*xi;
c2=(C13+C44)*1i.*xi;
b1=rho.*w.^2 - C11.*xi.^2;
b2=rho.*w.^2-C44.*xi.^2;
The vpasolve part ,i.e.,
syms xr
rsol=vpasolve((a1*xr+b1).*(a2*xr+b2)-c1*c2*xr==0,xr);
rsol=eval(rsol);
r=[-sqrt(rsol)];
r1=r(1,:);
r2=r(2,:);
can also be vectorized by using fsolve as it is shown here:
fun=#(Xr) (a1.*Xr+b1).*(a2.*Xr+b2)-c1.*c2.*Xr
x01=-ones(2*Nxi);
x02=ones(2*Nxi);
options.Algorithm= 'trust-region-reflective'
options.JacobPattern=speye(4*Nxi^2);
options.PrecondBandWidth=0;
[rsol1]=fsolve(fun,x01,options);
[rsol2]=fsolve(fun,x02,options);
The other part, i.e,
syms As1 As2;
try
[Ass1,Ass2]=vpasolve(BC11*As1+BC12*As2==0,BC21*As1+BC22*As2+Fdf1==0,As1,As2);
A1=eval(Ass1);
A2=eval(Ass2);
catch
A1=0.0;
A2=0.0;
end
since contains linear equations and in each row has two dependent equations can be solved as it is shown here:
funAB1=#(As1) BC11.*As1+BC12.*((-Fdf2-BC21.*As1)./BC22);
x0AB=ones(2*Nxi)+1i*ones(2*Nxi);
options.Algorithm= 'trust-region-reflective';
options.JacobPattern=speye(4*Nxi^2);
options.PrecondBandWidth=0;
[A1]=fsolve(funAB1,x0AB,options);
A2=((-Fdf2-BC21.*A1)./BC22);
However, both can also be solved analytically, i.e.,
AAr=a1.*a2;
BBr=a1.*b2+b1.*a2-c1.*c2;
CCr=b1.*b2;
Xr1=(-BBr+sqrt(BBr^.2-4.*AAr.*CCr))./(2.*AAr);
Xr2=(-BBr-sqrt(BBr^.2-4.*AAr.*CCr))./(2.*AAr);
r1=-sqrt(Xr1(:,:));
r2=-sqrt(Xr2(:,:));
I'm trying to differentiate WhittakerM function.
To solve the WhittakerM equation we have:
dsolve( 'D2y+(-1/4+Landa/r+(1/4-(L+1/2)^2)/r^2)*y=0' ,'r')
C1*WhittakerM(Landa,L+1/2,r)+C2*WhittakerW(Landa,L+1/2,r)
from the boundary condition I only need WhittakerM(Landa,L+1/2,r)/rthat 1/ris added for the condition of the problem. I'm trying to defferentiate it then substitute in some points, but there are some errorin subsand diff.
Landa=1;L=0; % # for simplicity
R1=inline('WhittakerM(Landa,L+1/2,r)/r','r');
Rp1=diff(R1,r);
r=1:0.01:20;
R1sub=eval(R1,r);
Rp1sub=eval(Rp1,r);
Do u have any idea?
Regardless of the errors above (honestly I do not fully understand what you are trying to achieve with the substitution), using the symbolic toolbox might be a good start here:
syms r
%defines a symbolic function
R1(r)=whittakerM(landa,l+1/2,r)/r
%differentiate
Rp1=diff(R1,r);
%evaluate
Rp1_e=Rp1(1:0.01:20)
The following command
syms x real;
f = #(x) log(x^2)*exp(-1/(x^2));
fp(x) = diff(f(x),x);
fpp(x) = diff(fp(x),x);
and
solve(fpp(x)>0,x,'Real',true)
return the result
solve([0.0 < (8.0*exp(-1.0/x^2))/x^4 - (2.0*exp(-1.0/x^2))/x^2 -
(6.0*log(x^2)*exp(-1.0/x^2))/x^4 + (4.0*log(x^2)*exp(-1.0/x^2))/x^6],
[x == RD_NINF..RD_INF])
which is not what I expect.
The first question: Is it possible to force Matlab's solve to return the set of all solutions?
(This is related to this question.) Moreover, when I try to solve the equation
solve(fpp(x)==0,x,'Real',true)
which returns
ans =
-1.5056100417680902125994180096313
I am not satisfied since all solutions are not returned (they are approximately -1.5056, 1.5056, -0.5663 and 0.5663 obtained from WolframAlpha).
I know that vpasolve with some initial guess can handle this. But, I have no idea how I can generally find initial guessed values to obtain all solutions, which is my second question.
Other solutions or suggestions for solving these problems are welcomed.
As I indicated in my comment above, sym/solve is primarily meant to solve for analytic solutions of equations. When this fails, it tries to find a numeric solution. Some equations can have an infinite number of numeric solutions (e.g., periodic equations), and thus, as per the documentation: "The numeric solver does not try to find all numeric solutions for [the] equation. Instead, it returns only the first solution that it finds."
However, one can access the features of MuPAD from within Matlab. MuPAD's numeric::solve function has several additional capabilities. In particular is the 'AllRealRoots' option. In your case:
syms x real;
f = #(x)log(x^2)*exp(-1/(x^2));
fp(x) = diff(f(x),x);
fpp(x) = diff(fp(x),x);
s = feval(symengine,'numeric::solve',fpp(x)==0,x,'AllRealRoots')
which returns
s =
[ -1.5056102995536617698689500437312, -0.56633904710786569620564475006904, 0.56633904710786569620564475006904, 1.5056102995536617698689500437312]
as well as a warning message.
My answer to this question provides other way that various MuPAD solvers can be used, particularly if you can isolate and bracket your roots.
The above is not going to directly help with your inequalities other than telling you where the function changes sign. For those you could try:
s = feval(symengine,'solve',fpp(x)>0,x,'Real')
which returns
s =
(Dom::Interval(0, Inf) union Dom::Interval(-Inf, 0)) intersect solve(0 < 2*log(x^2) - 3*x^2*log(x^2) + 4*x^2 - x^4, x, Real)
Try plotting this function along with fpp.
While this is not a bug per se, The MathWorks still might be interested in this difference in behavior and poor performance of sym/solve (and the underlying symobj::solvefull) relative to MuPAD's solve. File a bug report if you like. For the life of me I don't understand why they can't better unify these parts of Matlab. The separation makes not sense from the perspective of a user.
I have the following script that defines a function and sets up and equation:
H = #(f) sum(log(f));
f = rand(1, 1);
syms a
H(f)-H(f-a)
I want to solve H(f)-H(f-a)=0 for a. I tried using fzero in the following manner, fzero('H(f)-H(f-a)', 0), but this doesn't yield me anything useful.
well the clear answer for this would be that a=0. This is of course the obvious answer, and may be why you are not getting "anything useful", especially when only considering one value for f.
However, you can actually get useful information when you add more values and use the solve function.
H = #(f) sum(log(f));
f = rand(5, 1);
syms a
temp = H(f)-H(f-a);
out = double(solve(temp == 0,a));
This will output a vector with all values of a that are solutions to your function. Just note that this may not result in the best answers in general, numerical solving of high order functions is rather difficult and unreliable. Because of this, it may take a long time to run as well.
I have a Matlab project in which I need to make a GUI that receives two mathematical functions from the user. I then need to find their intersection point, and then plot the two functions.
So, I have several questions:
Do you know of any algorithm I can use to find the intersection point? Of course I prefer one to which I can already find a Matlab code for in the internet. Also, I prefer it wouldn't be the Newton-Raphson method.
I should point out I'm not allowed to use built in Matlab functions.
I'm having trouble plotting the functions. What I basically did is this:
fun_f = get(handles.Function_f,'string');
fun_g = get(handles.Function_g,'string');
cla % To clear axes when plotting new functions
ezplot(fun_f);
hold on
ezplot(fun_g);
axis ([-20 20 -10 10]);
The problem is that sometimes, the axes limits do not allow me to see the other function. This will happen, if, for example, I will have one function as log10(x) and the other as y=1, the y=1 will not be shown.
I have already tried using all the axis commands but to no avail. If I set the limits myself, the functions only exist in certain limits. I have no idea why.
3 . How do I display numbers in a static text? Or better yet, string with numbers?
I want to display something like x0 = [root1]; x1 = [root2]. The only solution I found was turning the roots I found into strings but I prefer not to.
As for the equation solver, this is the code I have so far. I know it is very amateurish but it seemed like the most "intuitive" way. Also keep in mind it is very very not finished (for example, it will show me only two solutions, I'm not so sure how to display multiple roots in one static text as they are strings, hence question #3).
function [Sol] = SolveEquation(handles)
fun_f = get(handles.Function_f,'string');
fun_g = get(handles.Function_g,'string');
f = inline(fun_f);
g = inline(fun_g);
i = 1;
Sol = 0;
for x = -10:0.1:10;
if (g(x) - f(x)) >= 0 && (g(x) - f(x)) < 0.01
Sol(i) = x;
i = i + 1;
end
end
solution1 = num2str(Sol(1));
solution2 = num2str(Sol(2));
set(handles.roots1,'string',solution1);
set(handles.roots2,'string',solution2);
The if condition is because the subtraction will never give me an absolute zero, and this seems to somewhat solve it, though it's really not perfect, sometimes it will give me more than two very similar solutions (e.g 1.9 and 2).
The range of x is arbitrary, chosen by me.
I know this is a long question, so I really appreciate your patience.
Thank you very much in advance!
Question 1
I think this is a more robust method for finding the roots given data at discrete points. Looking for when the difference between the functions changes sign, which corresponds to them crossing over.
S=sign(g(x)-f(x));
h=find(diff(S)~=0)
Sol=x(h);
If you can evaluate the function wherever you want there are more methods you can use, but it depends on the size of the domain and the accuracy you want as to what is best. For example, if you don't need a great deal of accurac, your f and g functions are simple to calculate, and you can't or don't want to use derivatives, you can get a more accurate root using the same idea as the first code snippet, but do it iteratively:
G=inline('sin(x)');
F=inline('1');
g=vectorize(G);
f=vectorize(F);
tol=1e-9;
tic()
x = -2*pi:.001:pi;
S=sign(g(x)-f(x));
h=find(diff(S)~=0); % Find where two lines cross over
Sol=zeros(size(h));
Err=zeros(size(h));
if ~isempty(h) % There are some cross-over points
for i=1:length(h) % For each point, improve the approximation
xN=x(h(i):h(i)+1);
err=1;
while(abs(err)>tol) % Iteratively improve aproximation
S=sign(g(xN)-f(xN));
hF=find(diff(S)~=0);
xN=xN(hF:hF+1);
[~,I]=min(abs(f(xN)-g(xN)));
xG=xN(I);
err=f(xG)-g(xG);
xN=linspace(xN(1),xN(2),15);
end
Sol(i)=xG;
Err(i)=f(xG)-g(xG);
end
else % No crossover points - lines could meet at tangents
[h,I]=findpeaks(-abs(g(x)-f(x)));
Sol=x(I(abs(f(x(I))-g(x(I)))<1e-5));
Err=f(Sol)-g(Sol)
end
% We also have to check each endpoint
if abs(f(x(end))-g(x(end)))<tol && abs(Sol(end)-x(end))>1e-12
Sol=[Sol x(end)];
Err=[Err g(x(end))-f(x(end))];
end
if abs(f(x(1))-g(x(1)))<tol && abs(Sol(1)-x(1))>1e-12
Sol=[x(1) Sol];
Err=[g(x(1))-f(x(1)) Err];
end
toc()
Sol
Err
This will "zoom" in to the region around each suspected root, and iteratively improve the accuracy. You can tweak the parameters to see whether they give better behaviour (the tolerance tol, the 15, number of new points to generate, could be higher probably).
Question 2
You would probably be best off avoiding ezplot, and using plot, which gives you greater control. You can vectorise inline functions so that you can evaluate them like anonymous functions, as I did in the previous code snippet, using
f=inline('x^2')
F=vectorize(f)
F(1:5)
and this should make plotting much easier:
plot(x,f(x),'b',Sol,f(Sol),'ro',x,g(x),'k',Sol,G(Sol),'ro')
Question 3
I'm not sure why you don't want to display your roots as strings, what's wrong with this:
text(xPos,yPos,['x0=' num2str(Sol(1))]);